Determine the value of $p>0$ for which $$\sum_{n=1}^{\infty}\frac{(-1)^{\lfloor{\sqrt{n}}\rfloor}}{n^p}$$ converges.

By considering $\lfloor{\sqrt{n}}\rfloor$, we see the series is

$$\sum_{k\ge1} (-1)^k\sum_{i=k^2}^{(k+1)^2-1}\frac1{i^p}$$


For $p>1$, the series converges absolutely.

For $p=1$: from the expression $$ \sum_{k\ge1} (-1)^k\sum_{i=k^2}^{(k+1)^2-1}\frac1{i^p} $$ we have the estimates $$ \log\frac{(k+1)^2}{k^2}=\int_{k^2}^{(k+1)^2}\frac{dt}t\,\leq\sum_{j=k^2}^{(k+1)^2-1}\frac1j\leq\int_{k^2-1}^{(k+1)^2-1}\,\frac{dt}t=\log\frac{(k+1)^2-1}{k^2-1}\\ =\log\frac{k(k+2)}{(k-1)(k+1)}. $$ This implies that $$ \lim_{k\to\infty}\sum_{j=k^2}^{(k+1)^2-1}\frac1j=0 $$ and, moreover, $$ \sum_{j=(k+1)^2}^{(k+2)^2-1}\frac1j\leq\log\frac{(k+1)(k+3)}{k(k+2)}\leq\log\frac{(k+1)^2}{k^2}\leq\sum_{j=1k^2}^{(k+1)^2-1}\frac1j $$ (the inequality in the middle is equivalent to $(k+3)/(k+2)\leq(k+1)/k$). Then Leibnitz's Criterion for Alternate Series implies that the series is convergent for $p=1$.

Edit: the second part of the proof was wrong. Here is a better argument.

Assume $p<1$.

We have the estimates, for $k\geq1/(2^{1-p}-1)$ (i.e. $2k+1\leq2^{1-p}k$), $$ \sum_{j=k^2}^{(k+1)^2-1}\frac1{j^p}\geq\frac{2k+1}{((k+1)^2-1)^p}=\frac{2k+1}{(k^2+2k)^p}\geq\frac{2k}{(k^2+2k)^p}\geq\frac{2k}{2^pk^{2p}}=2^{1-p}\,k^{1-2p}, $$ $$ \sum_{j=k^2}^{(k+1)^2-1}\frac1{j^p}\leq\frac{2k+1}{k^{2p}}\leq\frac{2^{1-p}k}{k^{2p}}=2^{1-p}k^{1-2p}. $$ The first estimate shows that the series diverges for $p\leq1/2$, as then the sums over $j$ are not even going to zero.

For $p>1/2$, we have $$ \sum_{k=1}^\infty(-1)^k\sum_{j=k^2}^{(k+1)^2-1}\frac1{j^p}=-\sum_{k=1}^\infty\left(\sum_{j=(2k-1)^2}^{(2k)^2-1}\frac1{j^p}-\sum_{j=(2k)^2}^{(2k+1)^2-1}\frac1{j^p}\right). $$ The series on the right converges absolutely. Indeed, for big enough $k$, $$ \sum_{j=(2k-1)^2}^{(2k)^2-1}\frac1{j^p}-\sum_{j=(2k)^2}^{(2k+1)^2-1}\frac1{j^p}\leq2^{1-p}(2k-1)^{1-2p}-2^{1-p}(2k)^{1-2p}=2^{1-p}(2k)^{1-2p}\left(\left(1-\frac1{2k}\right)^{1-2p}-1\right)=2^{1-p}(2k)^{1-2p}\left(-\frac{1-2p}{2k}+O(k^{-2})\right)=-2^{1-p}(1-2p)\left((2k)^{-2p}+O(k^{-1-2p})\right). $$ So the convergence of the series is given by the behaviour of $k^{-2p}$. For $p\leq1/2$, we have $2p\leq1$ and the series diverges. For $p>1/2$, $2p>1$ and the seires converges.

  • $\begingroup$ It's not true that $(k+3)/(k+2)\le(k+1)/k$... $\endgroup$ – user70520 Oct 18 '13 at 8:16
  • $\begingroup$ I don't think it's true in general that $\frac{(k+1)^{1-p}(k+3)^{1-p}-k^{1-p}(k+2)^{1-p}}{1-p} \leq\frac{(k+1)^{2(1-p)}-k^{2(1-p)}}{1-p}$; set $f(x)=x^{1-p}((x+2)^{1-p}-x^{1-p})$ and calculate $f'(x)$ (we want this to be less than $0$); then it boils down to showing $x+1\le(x+2)^{p}x^{1-p}$, which isn't true in general. $\endgroup$ – user70520 Oct 18 '13 at 8:32
  • $\begingroup$ Regarding your first comment, $k(k+3)=k^2+3k\leq k^2+3k+2=(k+1)(k+2)$. Your second comment is spot on: the inequality I wrote is false. $\endgroup$ – Martin Argerami Oct 18 '13 at 15:50
  • $\begingroup$ I have included a new argument. $\endgroup$ – Martin Argerami Oct 18 '13 at 20:55

The limsup and the liminf of the sequence of partial sums of the series are the limits of the partial sums at $(2k+1)^2-1$ and at $(2k)^2-1$ respectively hence the series converges if and only if $S_k\to0$ where the $k$th slab $S_k$ is $$ S_k=\sum_{n=k^2}^{(k+1)^2-1}\frac1{n^p}. $$ There are $2k+1$ terms in $S_k$, all between $1/k^{2p}$ and $1/(k+1)^{2p}$, hence $$ \frac{2k}{(k+1)^{2p}}\leqslant S_k\leqslant\frac{2k+1}{k^{2p}}. $$ This proves that $S_k=\Theta(k^{1-2p})$ hence the original series converges if and only if $$ p\gt\frac12. $$


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